Alg7.4

Prove that if $x^2=1\forall x\isin G\,$ , then $G\,$ is abelian.

$(1) \forall x \isin G, x\cdot x=1 \implies x=x^{-1}\,$

If $a,b\isin G, (a a)(b b)=1\,$

$a^{-1} a a b b b^{-1} = a^{-1} b^{-1}\,$

$a b = a^{-1} b^{-1} = (ba)^{-1}\,$

But from $(1), ba = (ba)^{-1}\,$

So now $a b = b a\,$

Or, even simpler:

$a b = 1 a b 1 = (b b) a b (a a) = b (b a) (b a) a = b 1 a = b a \,$

Since this proof doesn't use inverses, the property holds for all monoids, not only groups.

But every monoid, for which this property holds, is in fact a group.